Deriving 3 equations of motion (from v-t graph)

- [Instructor] Imagine an airplane is accelerating on this runway. And let's say, when the airplane is somewhere over here, we start our timer and let's say it has some velocity. That's what I've blurted to show that it's moving, it's not at rest. And let's say it's velocity at that point is u. And the question we wanna try and answer in this video is after sometime, let's say, t, where will this airplane be and what will be its new velocity? What I mean is if we wait for some time, t, we can kind of guess the airplane will be somewhere over here but it will be a little bit faster because it is accelerating. It's constantly increasing its velocity. So now the question is, what is the velocity at that point? And how far would it have traveled? And because questions like these are very common, let's build equations to calculate these things. And one of the ways to calculate them and build equations is by drawing a velocity-time graph. Look at this graph. It's also telling us the same story. When we start our timer, that's when time is zero, the velocity of the airplane is u. And after time, t, the new velocity of our airplane is v. And a very important thing to notice is because you've drawn a straight line, this means that the acceleration of our airplane is a constant. But what does that mean? Well, that just means you are assuming that the velocity of our airplane is increasing by the same amount every second. For example, if initially over here its velocity was say 20, next second it said becomes 25, then the next second, it should become 30. Again, increase by five. Next second it should become 35. It should always increase by five. That's what we mean by constant acceleration. So we'll go ahead and build equations for velocity and displacement for constant accelerations. All right, let's do this. Okay. Let's start with V. To calculate this, we just have to remember the definition of acceleration. Remember that acceleration is, change in velocity divided by time. It's a measure of how quickly the velocity is changing, right? So in our case, what's the change in velocity? Well, the final velocity is v, initial velocity is u, so the change is v minus u. V minus u divide by what is the time it takes to change that. From here to here, the time is t. And since we want to calculate what v is, all we have to do is rearrange this equation and do some algebra. So great idea to pause the video and see if you can try that yourself first. All right. If you have done that rearrangement and if you've done that algebra, then you will end up with v equals u plus at. And using this equation, we can calculate what v is, final velocity. Now, let me tell you what they do in your textbooks. The textbooks they derive this from the graph. This is what they do. If you look at the graph, this length over here is the same as this length which is v and this length is u. And we can write that u on this side. And so in the graph, look v must be u plus some thing over here. But what is that something? And then they use this equation and say look, v equals u plus at. So v should be u plus this something must be at from the equation. So this length should be at and tara, v should be u plus at. It's actually the same thing, right? So the important thing is this is not a new equation. It is just a rearrangement of the definition. So even if you forget this equation, it's fine as long as you remember what acceleration is. All right, let's do the next one. Next, we'll build an equation for displacement. Now can't I use displacement equals velocity into time? No, I can't because this equation is only useful provided velocity is a constant. In our example, our velocity is continuously changing. So what number would you substitute for velocity? So then what to do? Well in a previous video we had seen, that when you multiply velocity with time, graphically it represents the area. So in the velocity-time graph, the area under that graph gives us the displacement. So over here, all we have to do is calculate this area. And if you need more clarity on why this is so, we've talked about this in previous videos. So you can always go back and refer. But anyways, to calculate displacement, I just need to figure out what this area is. So I don't need this diagram anymore. Let's get rid of that and make some more room. All right. So, second derivation is displacement, s, is going to be the area under this graph. Now, how do we calculate that area? Well, if you look at this graph carefully, you can see two shapes, one rectangle and one triangle. So we can say, the area is going to be the area of the rectangle plus the area of this triangle. Now again, a good time to pause the video and see if you can try this yourself. Please try this because I'm sure you know how to calculate the areas of these two shapes. Go ahead, try this. All right. So the area of our rectangle, is this rectangle is going to be length into breath, is going to be u into t. So that's going to be ut plus what's the area of a triangle? Well that's half times base into height. So half into base, base is t into height of this triangle which is at. So that's going to be a into t and this gives us a into t squared. And so our equation becomes s equals ut plus half at squared. Two down, one more equation to go. At first you may be like, wait a second. Why do we need a third equation? I mean, we wanted v and s and we got that right? Well, you're right. Actually, we just need these two equations. However, if you see properly, you can see that both these equations depend on time. And sometimes in our problems, time will not be given to us. And it's for that reason just to make our calculations a little faster, it's a good idea to build a third equation which does not have time in it. So this will be purely mathematical, no more physics. Physics is done. All we'll do is use these two equations and eliminate time in it. All right? So what we can do is we will find out what t is from this equation and then we'll plug it in this equation and see what we get. Now, if you directly do this, the process might be little longer because there's a t here and there's a t squared over there so it might become big. Just to make it shorter, you know what I'll do? I will recalculate the area of this. This time, I'll calculate it as the area of the entire ship. I'll calculate as the area of this trapezium. You'll see why, you'll see why. So, do you remember what the area of the trapezium is? We calculate the area of trapezium in general as half into base into the sum of the two parallel heights. So let's do that. Let's see what we get. And you'll see why I'm doing it this way. So it's going to be half into base which is time, t, into the sum of the two parallel heights. These are the two parallel heights. So it's going to be u plus v, u plus v. Now look at this equation. This equation only has t in it. So it's a very nice equation to substitute the value of t. That's the only reason I am using this equation and not this one, okay. So now let's find what t is from here and substitute in this. So if you go over here, what is t equal to? Well if you look carefully, t is going to be, again if you just do a little bit of algebra, you'll see t equals v minus u divided by a. Can you see that? Again, if you need some time just pause the video in between and convince yourself of this. It's going to be v minus u divided by a. Now all you have to do is take this equation for time and plug it in over here. And again, it's one of those moments where you could pause the video and see if you can try this yourself, see what equation you end up with. All right, let's do this. Let me move this thing up because we don't need any of that anymore. We just need this. So if we substitute, we'll get s is half into time. The time is v minus u divide by a into u plus v. Now if you simplify this, let's see what we get. We get half into, look at the numerator. We have v minus u into v plus u. Ooh, that is v squared minus, minus u squared. And in the denominator, I just have an a. And again, if we do some algebra and we'll rearrange this, we will now end up with v squared minus u squared equals you get 2as. So we end up with 2as and that is our equation number three. As you can see, it is independent of time. So here it is, all three equations in one place. I know it looks a little crowded, sorry about that. But before we end, let's just see if these equations make sense to us. And one of the ways to do that is to check what happens if we put a equal to zero. Think about it. If a is zero, then that means there is no acceleration and it means our body is moving with a constant velocity. Let's see what happens to these equations and let's think whether that makes sense to us. In the first equation if you put a equal to zero, this whole term vanishes and that gives us v equals u. Now, does that make sense? Well, it does. Because if acceleration is zero, we would expect it's velocity to be a constant. That means, regardless of what time you check, it's final velocity must always be the same as its initial velocity. So that equation makes sense. Similarly, if you were to put a equal to zero in the second equation, this term vanishes and now we get s equals just ut. If we look carefully, we are going back to this now. Displacement equals velocity into time. Makes sense, right? Because there is no acceleration, this equation goes back to the old equation we used to use in uniform motion. Perfect. And similarly, if you put a equal to zero in the third equation, you'll again get v equals u which we already discussed. Also, instead of increasing its speed, if the objects were decreasing its speed, even then these equations work. In that case, all we have to do is put a negative value for the acceleration, deceleration and then these equations will work. So to summarize, how do we derive these equations? Well, the first equation is for v and that is done by just rearranging our acceleration definition. It's not a new equation. The second equation is for displacement, s and that is calculated as the area under the graph. So it's just the area of rectangle plus the area of the triangle. Two equations are actually enough but we build a third equation which is independent of time. And to do that, we can actually take t from here and substitute over here. To minimize the number of steps, I recalculated the displacement as the area of trapezium so that we just have one t and then substitute it. But even if you directly substitute it over here, you end up with the same answer. You'll just have two or three additional steps. And yeah, one more, almost forgot. These equations only work provided our acceleration is a constant. If the acceleration is changing, then these equations don't work. Now we're done.